Question

When 81.0 g of aluminium is allowed to react with 128.0 g of oxygen gas, the mass of aluminium oxide produced in grams is ----- (nearest integer).

Hint:

Use stoichiometry to calculate the mass of the product based on the limiting reagent in the reaction.

Answer 1

Correct Answer: 153

To determine the mass of aluminium oxide (Al₂O₃) produced, the balanced chemical equation is:

4Al + 3O₂ → 2Al₂O₃

1. Molar Masses: 
- Al = 27 g/mol
- O₂ = 32 g/mol
- Al₂O₃ = (2×27) + (3×16) = 102 g/mol

2. Limiting Reagent Determination:

Moles of Al in 81.0 g:

n(Al) = 81.0 g / 27 g/mol = 3.0 mol

Moles of O₂ in 128.0 g:

n(O₂) = 128.0 g / 32 g/mol = 4.0 mol

The stoichiometry indicates 4 moles of Al react with 3 moles of O₂. Calculate the oxygen needed for 3.0 mol of Al:

Required O₂ = 3.0 mol Al × (3 mol O₂ / 4 mol Al) = 2.25 mol O₂

Since 2.25 mol of O₂ is less than the available 4.0 mol, Al is the limiting reagent.

3. Mass of Al₂O₃ Calculation:

From 3.0 mol of Al, the moles of Al₂O₃ produced are:

n(Al₂O₃) = 3.0 mol Al × (2 mol Al₂O₃ / 4 mol Al) = 1.5 mol

Mass of Al₂O₃ = 1.5 mol × 102 g/mol = 153.0 g

The mass of aluminium oxide produced is 153 g.

This result is within the expected range of 153 to 153 g.